LeetCode.394 Decode String

题目：

Given an encoded string, return it's decoded string.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.

You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.

Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won't be input like 3a or 2[4].

Examples:

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

分析：

class Solution {
    public String decodeString(String s) {
        // 给定一个加密字符串（有效，没有其他额外空间。该加密字符串一定是满足下面加密规则的字符串。），返回其解密后的字符串
		// 加密规则：k[]外面的为加密k次，里面为加密数字。k一定是一个正数
		// 思路：使用两个stack实现（一个用来存储数字，一个用来存储对应的字符串），每判断到一个]，对两个stack的元素做对应的处理。
		// 处理完后，将字符接在前一个字符串的末尾，判断完直接返回stack.pop()
		String res = "";
		Stack<Integer> numStack = new Stack<Integer>();
		Stack<String> strStack = new Stack<String>();

		int index = 0;
		while (index < s.length()) {
			if (Character.isDigit(s.charAt(index))) {
				// 数字
				int count = 0;
				while (Character.isDigit(s.charAt(index))) {
					count = count * 10 + (s.charAt(index) - '0');
					index++;
				}

				// 将数字压入栈
				numStack.push(count);
			} else if (s.charAt(index) == '[') {
				strStack.push(res);
				res = "";
				index++;
			} else if (s.charAt(index) == ']') {
				StringBuilder sb = new StringBuilder(strStack.pop());
				int times = numStack.pop();
				for (int j = 1; j <= times; j++) {
					sb.append(res);
				}
				res = sb.toString();
				index++;
			} else {
				// 直接添加的字符串
				res += s.charAt(index++);
			}
		}

		return res;
    }
}